# Bonus work: 3 Course: M339W/M389W - Financial Math for Actuaries

```Bonus work: 3
Course: M339W/M389W - Financial Math for Actuaries
Page: 1 of 2
University of Texas at Austin
Quiz # 3
Log-normal stock prices.
Problem 3.1. (5 pts) A non-dividend-paying stock is valued at \$100.00 per share. The annual expected
(rate of) return is 12.0% and the standard deviation of annualized returns is given to be 0.30. If the stock
price is modeled using the lognormal distribution (as discussed in class), what is the probability that the
time−2 stock price exceeds \$95?
(e) None of the above.
Solution: (d)
In our usual notation, it is given that S(0) = 100, α = 0.12 and σ = 0.30. As we have learned in class, the
required probability can be expressed as
P[S(2) > 95] = N (dˆ2 )
with N denoting the standard normal cumulative distribution function and
S(0)
1
1
+ (α − σ 2 ) · 2 = 0.474453.
dˆ2 = √ ln
95
2
σ 2
Using the standard normal tables, we get N (dˆ2 ) = 0.682411.
Problem 3.2. (5 points) A non-dividend-paying stock is valued at \$100.00 per share. The annual expected
(rate of) return is 12.0% and the standard deviation of annualized returns is given to be 0.30. If the stock
price is modeled using the lognormal distribution (as discussed in class), what is the following conditional
expectation?
E[S(2) | S(2) > 95]
(a)
(b)
(c)
(d)
(e)
None of the above.
Solution: (d)
In our usual notation, it is given that S(0) = 100, α = 0.12 and σ = 0.30. As we have learned in class, the
conditional expectation from above can be expressed as
N (dˆ1 )
E[S(2) | S(2) > 95] = S(0)e2α
.
N (dˆ2 )
with N denoting the standard normal cumulative distribution function and
1
1
S(0)
+ (α + σ 2 ) · 2 = 0.898717,
dˆ1 = √ ln
95
2
σ 2
√
ˆ
ˆ
d2 = d1 − σ 2 = 0.474453.
Using the standard normal tables, we get N (dˆ1 ) = 0.815598 and N (dˆ2 ) = 0.682411. So,
E[S(2) | S(2) > 95] = 151.936.
ˇ
Instructor: Milica Cudina
Bonus work: 3
Course: M339W/M389W - Financial Math for Actuaries
Page: 2 of 2
Problem 3.3. (10 points) A non-dividend-paying stock is valued at \$75.00 per share. The annual expected
(rate of) return is 16.0% and the standard deviation of annualized returns is given to be 0.30. If the stock
price is modeled using the lognormal distribution (as discussed in class), what is the constant sU
1/2 such that
P[S(1/2) > sU
1/2 ] ≤ 0.05.
Solution: Note that the 95th percentile of the standard normal distribution equals 1.645. So,
(0.16− 12 ×0.32 )× 12 +0.3× √12 ×1.645
sU
1/2 = 75e
= 112.61.
Problem 3.4. (10 points) A continuous-dividend-paying stock is valued at \$75.00 per share. Its dividend
yield is 0.03. The time−t realized (rate of) return is modeled as
R(0, t) ∼ N (mean = 0.035t, variance = 0.09t)
Find the probability that the time−4 stock price exceeds today’s stock price.
Solution: We need to find
P[S(4) > S(0)]
with
S(t) = S(0)eR(0,t) .
Since R(0, t) follows the normal distribution with the above parameters, we have
0.035 × 4
R(0,4)
P[S(4) > S(0)] = P[S(0)e
> S(0)] = P[R(0, 4) > 0] = 1 − N −
= N (0.23) = 0.591.
0.3 × 2
Problem 3.5. (10 points) A continuous-dividend-paying stock is valued at \$75.00 per share. Its dividend
yield is 0.03. The time−t realized (rate of) return is modeled as
R(0, t) ∼ N (mean = 0.035t, variance = 0.09t)
An investor purchases a single share of stock at time−0 and continuously and immediately reinvests any
dividends received in the same asset. What are the mean and median values of the investor’s position at
time−4?
Solution: The expected rate of return (per annum) is
1
0.035 + × 0.09 = 0.035 + 0.045 = 0.08.
2
The mean is
e4δ E[S(T )] = 75e4α = 75e0.32 = 103.285.
Similarly, the median is
77e4×0.035 = 75 × 1.15027 = 86.2705.
ˇ
Instructor: Milica Cudina
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